Question

Show that the relation $R$ in the set $Z$ of integers given by
$R=\left\{\right(a,b):2dividesa-b\}$

Answer 1

$2\left\{\right(a,b)=2dividesa=b\}$

where R is in set 2 of integers

a-a=0

2 divides a-a

$\Rightarrow (a,a)ϵR$ $\therefore R$ is reflexive

Let $(a,a)ϵR$

$\therefore 2$ divides $a-b\Rightarrow a-b=2n$

for same $nϵz\Rightarrow b-a=2(-n)$

$\Rightarrow$ 2 divides $b-a=(b,a)ϵR$

R is symmetric

Let (a,b) & $(b,c)ϵR$

2 divides a-b & b-c

$\therefore a-b=2n$ & $b-c=2n_2$

for same $n_1,n_2ϵ2$

$=2n_1+2n_2=a-c=2(n_1+n_2)$

$\therefore 2$ divides a-c

$(a,c)ϵR$

$\therefore (a,b)(b,c)ϵR\Rightarrow (a,c)ϵR$

$\therefore R$ is transitive

So, R is an equivalence relation